A motor shaft rotates at ω_1 = 6000 rpm . During braking…

A motor shaft rotates at $\omega_1 = 6000\ \mathrm{rpm}$. During braking, its angular velocity drops uniformly to $\omega_2 = 4800\ \mathrm{rpm}$ in $t = 4\ \mathrm{s}$. Find the angular acceleration of the shaft and the number of revolutions it makes during braking.

Explanation of Method

Braking a motor shaft: angular acceleration and revolutions.

Step 1: Convert the units

$1\ \mathrm{rpm} = \tfrac{2\pi}{60}\ \mathrm{rad/s}$: $\omega_1 = 6000 \cdot \frac{2\pi}{60} = 200\pi \approx 628.3\ \mathrm{rad/s}$.

\[ \omega_2 = 160\pi \approx 502.7\ \mathrm{rad/s}. \]

Step 2: Angular acceleration

\[ \alpha = \frac{\omega_2 - \omega_1}{t} = \frac{160\pi - 200\pi}{4} = -10\pi\ \mathrm{rad/s^2}. \]

\[\alpha = -10\pi \approx -31.4\ \mathrm{rad/s^2}\]

Step 3: Number of revolutions

With uniform deceleration, the average rate is the mean of the start and end rates. In revolutions per second (100 and 80 rev/s): $N = \frac{\nu_1 + \nu_2}{2}\,t = \frac{100 + 80}{2} \cdot 4$.

\[N = 360 \ \text{revolutions}\]

Check in radians

$ \theta = \omega_1 t + \tfrac12 \alpha t^2 = 800\pi - 80\pi = 720\pi\ \mathrm{rad} $, and $ \tfrac{720\pi}{2\pi} = 360\ \checkmark $

$\omega\ (\mathrm{rev/s})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $3$ $4$ $80$ $100$ $\text{area} = \tfrac{100 + 80}{2}\cdot 4 = 360\ \mathrm{rev}$ $\text{slope} = -5\ \mathrm{rev/s^2} = -10\pi\ \mathrm{rad/s^2}$

Angular velocity against time, in revolutions per second (6000 rpm = 100 rev/s, 4800 rpm = 80 rev/s). The area under the line is the number of revolutions; the slope is the angular acceleration.

Result

\[\boxed{\,\alpha = -10\pi \approx -31.4\ \mathrm{rad/s^2}\,}\]

\[\boxed{\,N = 360 \ \text{revolutions}\,}\]